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\begin{document}

\title{Complex Analysis}
\subtitle{Chapter 3. Analytic Functions as Mappings \\ Section 3. Linear Transformations }
%\institute{SLUC}
\author{LVA}
%\date
%\renewcommand{\today}{\number\year \,年 \number\month \,月 \number\day \,日}
%\date{ {2023年9月21日} }

\maketitle

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% 设置方程编号从16开始
\setcounter{equation}{6} % 因为设置为6，下一个使用的标签就会从7开始


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\begin{frame}{Contents 1-2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.] Elementary Point Set Topology
\begin{enumerate}
\item[1.1.] Sets and Elements
\item[1.2.] Metric Spaces
\item[1.3.] Connectedness
\item[1.4.] Compactness
\item[1.5.] Continuous Functions
\item[1.6.] Topological Spaces
\end{enumerate}

\item[2.] Conformality
\begin{enumerate}
\item[2.1.] Arcs and Closed Curves
\item[2.2.] Analytic Functions in Regions
\item[2.3.] Conformal Mapping
\item[2.4.] Length and Area
\end{enumerate}

\end{enumerate}

\end{frame}

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\begin{frame}{Contents 3-4}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}
\item[3.] {\color{red}Linear Transformations}
\begin{enumerate}
\item[3.1.] {\color{red}The Linear Group}
\item[3.2.] {\color{red}The Cross Ratio}
\item[3.3.] {\color{red}Symmetry}
\item[3.4.] {\color{red}Oriented Circles}
\item[3.5.] {\color{red}Families of Circles}
\end{enumerate}

\item[4.] Elementary Conformal Mappings
\begin{enumerate}
\item[4.1.] The Use of Level Curves
\item[4.2.] A Survey of Elementary Mappings
\item[4.3.] Elementary Riemann Surfaces
\end{enumerate}
 
\end{enumerate}

\end{frame}

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\begin{frame}{3. Linear Transformations. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.] 
{\color{red}Of all analytic functions the {\color{blue}first-order rational functions} have the simplest mapping properties, for they define mappings of the extended plane onto itself which are at the same time conformal and topological. 
}

\item[2.] 
The {\color{blue}linear transformations} have also very remarkable geometric properties, and for that reason their importance goes far beyond serving as simple examples of conformal mappings. 

\item[3.] 
The reader will do well to pay particular attention to this geometric aspect, for it will equip him with simple but very valuable techniques.

\end{enumerate}

\end{frame}

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\begin{frame}{3.1. The Linear Group. \hfill 作业3C-1 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.] 
We have already remarked in Chap. 2, Sec. 1.4 that a linear fractional transformation is 
\begin{equation}
w = S(z) = \frac{az+b}{cz+d}.
\label{eq-7}
\end{equation}
with $ad - bc \neq 0$ has an inverse
\begin{equation*}
z = S^{-1}(w) = \frac{dw-b}{-cw+a}.
%\label{eq-}
\end{equation*}

\item[2.] 
The special values $S(\infty) = a/c$ and $S(-d/c) = \infty$ can be introduced either by convention or as limits for $z\to\infty$ and $z\to -d/c$.

\item[3.] 
{\color{red}With the latter interpretation it becomes obvious that $S$ is a {\color{blue}topological mapping} of the extended plane onto itself, the topology being defined by distances on the Riemann sphere.
}

\end{enumerate}

\end{frame}

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\begin{frame}{3.1. The Linear Group. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[4.] 
For linear transformations we shall usually replace the notation $S(z)$ by $Sz$.

\item[5.] 
The representation (\ref{eq-7}) is said to be {\color{blue}normalized} if $ad-bc=1$.

\item[6.] 
It is clear that every linear transformation has two normalized representations, obtained from each other by changing the signs of the coefficients.

\end{enumerate}

\end{frame}

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\begin{frame}{3.1. The Linear Group. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}



\item[7.] 
A convenient way to express the linear transformation by use of homogeneous coordinates.

\item[8.] 
If we write $z = z_1/z_2$, $w = w_1/w_2$ we find that $w = Sz$ if
\begin{equation}
\begin{aligned}
w_1 &= az_1 + bz_2 \\
w_2 &= cz_1 + dz_2
\end{aligned}
\label{eq-8}
\end{equation}
or, in matrix notation, 
\begin{equation*}
\begin{aligned}
\begin{pmatrix} w_1 \\ w_2 \end{pmatrix} = 
\begin{pmatrix} a & b \\ c & d \end{pmatrix} 
\begin{pmatrix} z_1 \\ z_2 \end{pmatrix}.  
\end{aligned}
\end{equation*}

\end{enumerate}

\end{frame}

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\begin{frame}{3.1. The Linear Group. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[9.] 
{\color{red}The main advantage of this notation is that it leads to a simple determination of a composite transformation $w = S_1S_2z$. }

\item[10.] 
If we use subscripts to distinguish between the matrices that correspond to $S_1$, $S_2$ it is immediate that $S_lS_2$ belongs to the matrix product 
\begin{equation*}
\begin{pmatrix} a_1 & b_1 \\ c_1 & d_1 \end{pmatrix} 
\begin{pmatrix} a_2 & b_2 \\ c_2 & d_2 \end{pmatrix} 
=\begin{pmatrix} a_1a_2 + b_1c_2 & a_1b_2+b_1d_2 \\ c_1a_2+d_1c_2 & c_1b_2+d_1d_2 \end{pmatrix}. 
\end{equation*}


\end{enumerate}

\end{frame}

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\begin{frame}{3.1. The Linear Group. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[11.] 
{\color{red}All linear transformations form a group. }

\item[12.] 
Indeed, the associative law $(S_1S_2)S_3 = S_1(S_2S_3)$ holds for arbitrary transformations, the identity $w = z$ is a linear transformation, and the inverse of a linear transformation is linear.

\item[13.] 
The ratios $z_1: z_2 \neq 0:0$ are the points of the complex projective line, and (\ref{eq-8}) identifies the group of linear transformations with the one-dimensional projective group over the complex numbers, usually denoted by $P(1,\mathbb{C})$. 

\item[14.] 
If we use only normalized representations, we can also identify it with the group of two-by-two matrices with determinant 1 (denoted $SL(2,\mathbb{C}))$, except that there are two opposite matrices corresponding to the same linear transformation.

\end{enumerate}

\end{frame}

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\begin{frame}{3.1. The Linear Group. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}



\item[15.] 
We shall make no further use of the matrix notation, except for remarking that the simplest linear transformations belong to matrices of the form
\begin{equation*}
\begin{pmatrix} 1&\alpha \\ 0&1  \end{pmatrix}, \hspace{0.5cm}
\begin{pmatrix} k&0 \\ 0&1  \end{pmatrix}, \hspace{0.5cm}
\begin{pmatrix} 0&1 \\ 1&0  \end{pmatrix}.  
\end{equation*}

\item[16.] 
The first of these, $w = z + a$, is called a {\color{blue}parallel translation}. 
 
\item[17.] 
The second, $w = kz$, is a {\color{blue}rotation} if $|k| = 1$ and a {\color{blue}homothetic transformation} if $k > 0$.

\item[18.] 
The third transformation, $w = 1/z$, is called an {\color{blue}inversion}. 

\end{enumerate}

\end{frame}

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\begin{frame}{3.1. The Linear Group. \hfill 作业3C-2 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[19.] 
{\color{red}
If $c \neq 0$ we can write
\begin{equation*}
\frac{az+b}{cz+d} = \frac{bc-ad}{c^2(z+d/c)} + \frac{a}{c},
\end{equation*}
and this decomposition shows that the most general linear transformaion is composed by a translation, an inversion, a rotation, and a homothetic transformation followed by another translation. 
}

\item[20.] 
If $c = 0$, the inversion falls out and the last translation is not needed.

\end{enumerate}

\end{frame}

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\begin{frame}{3.1. The Linear Group. Exercise - 1}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Prove that the reflection $z \to \bar{z}$ is not a linear transformation. 
}

\item  



\end{itemize}

\end{frame}

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\begin{frame}{3.1. The Linear Group. Exercise - 2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If
$$
T_1z = \frac{z+2}{z+3},\hspace{0.5cm}
T_2z = \frac{z}{z+1},
$$
find $T_1T_2z$, $T_2T_1z$ and $T_1^{-1}T_2z$.
}

\item  


\end{itemize}

\end{frame}

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\begin{frame}{3.1. The Linear Group. Exercise - 3}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Prove that the most general transformation which leaves the origin fixed and preserves all distances is either a rotation or a rotation followed by reflexion in the real axis.
}

\item  


\end{itemize}

\end{frame}

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\begin{frame}{3.1. The Linear Group. Exercise - 4}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Show that any linear transformation which transforms the real axis into itself can be written with real coefficients.
}

\item  


\end{itemize}

\end{frame}

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\begin{frame}{3.2. The Cross Ratio. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.] 
Given three distinct points $z_2, z_3, z_4$ in the extended plane, there exists a {\color{blue}linear transformation} $S$ which carries them into $1, 0, \infty$ in this order. 

\item[2.] 
If none of the points is $\infty$, $S$ will be given by
\begin{equation}
Sz = \frac{z-z_3}{z-z_4} : \frac{z_2-z_3}{z_2-z_4}.
\label{eq-9}
\end{equation}

\item[3.] 
If $z_2,z_3$, or $z_4 = \infty$, the transformation reduces to
\begin{equation*}
\frac{z-z_3}{z-z_4}, \hspace{0.5cm} 
\frac{z_2-z_3}{z-z_4}, \hspace{0.5cm} 
\frac{z-z_3}{z_2-z_3}. 
%\label{eq-}
\end{equation*}
respectively.

\end{enumerate}

\end{frame}

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\begin{frame}{3.2. The Cross Ratio. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[4.] 
If $T$ were another linear transformation with the same property, then $ST^{-1}$ would leave $1, 0, \infty$ invariant. 

\item[5.] 
Direct calculation shows that this is true only for the identity transformation, and we would have $S = T$. 

\item[6.] 
We conclude that $S$ is uniquely determined. 

\item[7.] 
{\color{red}Definition 12. The cross ratio $(z_1,z_2,z_3,z_4)$ is the image of $z_1$ under the linear transformation which carries $z_2,z_3,z_4$ into $1, 0, \infty$. }

\end{enumerate}

\end{frame}

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\begin{frame}{3.2. The Cross Ratio. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[8.] 
The definition is meaningful only if $z_2,z_3,z_4$ are distinct. 

\item[9.] 
A conventional value can be introduced as soon as any three of the points are distinct, but this is unimportant. 

\item[10.] 
The cross ratio is invariant under linear transformations. 

\item[11.] 
In more precise formulation:

{\color{red}Theorem 12. If $z_1, z_2, z_3, z_4$ are distinct points in the extended plane and $T$ any linear transformation, then $(Tz_1,Tz_2,Tz_3,Tz_4) = (z_1,z_2,z_3,z_4)$. } 

\end{enumerate}

\end{frame}

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\begin{frame}{3.2. The Cross Ratio. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[12.] 
The proof is immediate, for if $Sz = (z,z_2,z_3,z_4)$, then $ST^{-1}$ carries $Tz_2, Tz_3, Tz_4$ into $1, 0, \infty$. 

\item[13.] 
By definition we have hence
\begin{equation*}
(Tz_1,Tz_2,Tz_3,Tz_4) = ST^{-1}(Tz_1) = Sz_1 = (z_1,z_2,z_3,z_4).
\end{equation*}

\item[14.] 
{\color{red}
With the help of this property we can immediately write down the linear transformation which carries three given points $z_1, z_2, z_3$ to prescribed positions $w_1, w_2, w_3$. 
}

\item[15.] 
{\color{red}
The correspondence must indeed be given by 
\begin{equation*}
(w, w_1,w_2,w_3) = (z,z_1,z_2,z_3).
\end{equation*}
}

\end{enumerate}

\end{frame}

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\begin{frame}{3.2. The Cross Ratio. \hfill 作业3C-3 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[16.] 
In general it is of course necessary to solve this equation with respect to $w$.

\item[17.] 
{\color{red}Theorem 13. The cross ratio $(z_1,z_2,z_3,z_4)$ is real if and only if the four points lie on a circle or on a straight line. }

\item[18.] 
This is evident by elementary geometry, for we obtain 
\begin{equation*}
\mathrm{arg}\, (z_1,z_2,z_3,z_4) = \mathrm{arg}\, \frac{z_1-z_3}{z_1-z_4} - \mathrm{arg}\, \frac{z_2-z_3}{z_2-z_4},
\end{equation*}
and if the points lie on a circle this difference of angles is either $0$ or $\pm \pi$, depending on the relative location. 

\item[19.] 
For an analytic proof we need only show that the image of the real axis under any linear transformation is either a circle or a straight line. 

\end{enumerate}

\end{frame}

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\begin{frame}{3.2. The Cross Ratio. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[20.] 
Indeed, $Tz = (z,z_2,z_3,z_4)$ is real on the image of the real axis under the transformation $T^{-1}$ and nowhere else. 

\item[21.] 
The values of $w = T^{-1}z$ for real $z$ satisfy the equation $Tw = \overline{Tw}$.

\item[22.] 
Explicitly, this condition is of the form
\begin{equation*}
\frac{aw+b}{cw+d} = \frac{\bar{a}\bar{w}+\bar{b}}{\bar{c}\bar{w}+\bar{d}}. 
\end{equation*}

\item[23.] 
By cross multiplication we obtain
\begin{equation*}
(a\bar{c}-c\bar{a})|w^2| + (a\bar{d}-c\bar{b})w + (b\bar{c}-d\bar{a})\bar{w} + 
b\bar{d}-d\bar{b} = 0.
\end{equation*}

\end{enumerate}

\end{frame}

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\begin{frame}{3.2. The Cross Ratio. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[24.] 
If $a\bar{c}-c\bar{a}=0$ this is the equation of a straight line, for under this condition the coefficient $a\bar{d}-c\bar{b}$ cannot also vanish. 

\item[25.] 
If $a\bar{c}-c\bar{a} \neq 0$ we can divide by this coefficient and complete the square.

\item[26.] 
After a simple computation we obtain
\begin{equation*}
\left\vert w + \frac{\bar{a}d-\bar{c}b}{\bar{a}c-\bar{c}a} \right\vert =
 \left\vert\frac{ad-bc}{\bar{a}c-\bar{c}a}\right\vert
\end{equation*}
which is the equation of a circle.

\item[27.] 
The last result makes it clear that we should not, in the theory of linear transformations, distinguish between circles and straight lines.  

\end{enumerate}

\end{frame}

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\begin{frame}{3.2. The Cross Ratio. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[28.] 
A further justification was found in the fact that both correspond to circles on the Riemann sphere. 

\item[29.] 
Accordingly we shall agree to use the word circle in this wider sense. 

\item[30.] 
The following is an immediate corollary of Theorems 12 and 13:

{\color{red}Theorem 14. A linear transformation carries circles into circles. }

\end{enumerate}

\end{frame}

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\begin{frame}{3.2. The Cross Ratio. Exercise 1. \hfill 作业3C-4 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Find the linear transformation which carries $0, i, -i$ into $1, -1, 0$.
}

\item  


\end{itemize}

\end{frame}

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\begin{frame}{3.2. The Cross Ratio. Exercise 2. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Express the cross ratios corresponding to the 24 permutations of four points in terms of $\lambda = (z_1,z_2,z_3,z_4)$.
}

\item  


\end{itemize}

\end{frame}

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\begin{frame}{3.2. The Cross Ratio. Exercise 3. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If the consecutive vertices $z_1, z_2, z_3, z_4$ of a quadrilateral lie on a circle, prove that 
$$
|z_1-z_3|\cdot |z_2-z_4|=
|z_1-z_2|\cdot |z_3-z_4| + |z_2-z_3|\cdot |z_1-z_4|
$$
and interpret the result geometrically.
}

\item  


\end{itemize}

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\begin{frame}{3.2. The Cross Ratio. Exercise 4. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Show that any four distinct points can be carried by a linear
transformation to positions $1$, $-1$, $k$, $-k$, where the value of $k$ depends on
the points.
How many solutions are there, and how are they related?
}

\item  


\end{itemize}

\end{frame}

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\begin{frame}{3.3. Symmetry. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.] 
{\color{red}The points $z$ and $\bar{z}$ are symmetric with respect to the real axis. }

\item[2.] 
A linear transformation with real coefficients carries the real axis into itself and $z$, $\bar{z}$ into points which are again symmetric. 

\item[3.] 
{\color{blue}More generally, if a linear transformation $T$ carries the real axis into a circle $C$, we shall say that the points $w = Tz$ and $w^* = T\bar{z}$ are symmetric with respect to $C$.}

\item[4.] 
This is a relation between $w$, $w^*$ and $C$ which does not depend on $T$.

\item[5.] 
For if $S$ is another transformation which carries the real axis into $C$, then $S^{-1}T$ is a real transformation, and hence $S^{-1}w = S^{-1}Tz$ and $S^{-1}w^* = S^{-1}T\bar{z}$ are also conjugate. 


\end{enumerate}

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\begin{frame}{3.3. Symmetry. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[6.] 
Symmetry can thus be defined in the following terms. 

\item[7.] 
{\color{red}Definition 13. The points $z$ and $z^*$ are said to be symmetric with respect to the circle $C$ through $z_1$, $z_2$, $z_3$ if and only if 
$$(z^*,z_1,z_2,z_3) = (z,z_1,z_2,z_3). $$ } 

\item[8.] 
The points on $C$, and only those, are symmetric to themselves. 

\item[9.] 
The mapping which carries $z$ into $z^*$ is a one-to-one correspondence and is 
called {\color{blue}reflection with respect to $C$}. 

\item[10.] 
Two reflections will evidently result in a linear transformation. 

\end{enumerate}

\end{frame}

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\begin{frame}{3.3. Symmetry. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[11.] 
{\color{red}We wish to investigate the geometric significance of symmetry. } 

\item[12.] 
Suppose first that $C$ is a straight line. 

\item[13.] 
Then we can choose $z_3 = \infty$ and the condition for symmetry becomes
\begin{equation}
\frac{z^*-z_2}{z_1-z_2} = \frac{\bar{z}-\bar{z_2}}{\bar{z}_1-\bar{z_2}}
\label{eq-10}
\end{equation}

\item[14.] 
Taking absolute values we obtain $|z^*-z_2| = |z-z_2|$.

\item[15.] 
Here $z_2$ can be any finite point on $C$, and we conclude that $z$ and $z^*$ are equidistant from all points on $C$.

\end{enumerate}

\end{frame}

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\begin{frame}{3.3. Symmetry. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[16.] 
By (\ref{eq-10}) we have further 
\begin{equation*}
\mathrm{Im}\,\frac{z^*-z_2}{z_1-z_2} = -\mathrm{Im}\,\frac{z-z_2}{z_1-z_2},
%\label{eq-10}
\end{equation*}
and hence $z$ and $z^*$ are in different half planes determined by $C$. 

\item[17.] 
We leave to the reader to prove that $C$ is the {\color{blue}bisecting normal} of the segment between $z$ and $z^*$.

\item[18.] 
Consider now the case of a finite circle $C$ of center $a$ and radius $R$.

\end{enumerate}

\end{frame}

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\begin{frame}{3.3. Symmetry. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[19.] 
Systematic use of the invariance of the cross ratio allows us to conclude as follows:
\begin{equation*}
\begin{aligned}
  & \overline{(z,z_1,z_2,z_3)} \\
= & \overline{(z-a,z_1-a,z_2-a,z_3-a)}  \\
= & \left( \bar{z}-\bar{a}, \frac{R^2}{z_1-a}, \frac{R^2}{z_2-a}, \frac{R^2}{z_3-a}\right)  \\
= & \left( \frac{R^2}{\bar{z}-\bar{a}},z_1-a,z_2-a,z_3-a \right)  \\
= & \left( \frac{R^2}{\bar{z}-\bar{a}}+a, z_1,z_2,z_3 \right). 
\end{aligned}
%\label{eq-10}
\end{equation*}

\end{enumerate}

\end{frame}

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\begin{frame}{3.3. Symmetry. \hfill 作业3C-5 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[20.] 
This equation shows that the symmetric point of $z$ is 
$$z^* = R^2/(\bar{z}-\bar{a}) + a $$
or that $z$ and $z^*$ satisfy the relation 
\begin{equation}
(z^*-a)(\bar{z}-\bar{a}) = R^2. 
\label{eq-11}
\end{equation}

\item[21.] 
The product $|z^*-a|\cdot |\bar{z}-\bar{a}|$ of the distances to the center is hence $R^2$. 

\item[22.] 
Further, the ratio $(z^*-a)/(z-a)$ is positive, which means that $z$ and $z^*$ are situated on the same half line from $a$.

\item[23.] 
{\color{red}There is a simple geometric construction for the symmetric point of $z$ (Fig. 3-2).} 

\item[24.] 
We note that the symmetric point of $a$ is $\infty$.

\end{enumerate}

\end{frame}

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\begin{frame}{3.3. Symmetry. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}


\begin{figure}[ht!]
\centering
\includegraphics[height=0.6\textheight, width=0.75\textwidth]{figure-3-2.png}
%\caption{Fig. 3-1. Doubly covered region }
\end{figure}

\end{frame}

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\begin{frame}{3.3. Symmetry. \hfill 作业3C-6 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[25.] 
{\color{red}Theorem 15. (The symmetry principle.) If a linear transformation carries a circle $C_1$ into a circle $C_2$, then it transforms any pair of symmetric points with respect to $C_1$ into a pair of symmetric points with respect to $C_2$.  }

\item[26.] 
Briefly, linear transformations preserve symmetry. 

\item[27.] 
If $C_1$ or $C_2$ is the real axis, the principle follows from the definition of symmetry. 

\item[28.] 
In the general case the assertion follows by use of an intermediate transformation which carries $C_1$ into the real axis. 


\end{enumerate}

\end{frame}

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\begin{frame}{3.3. Symmetry. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[29.] 
{\color{red}There are two ways in which the principle of symmetry can be used. }

\item[30.] 
If the images of $z$ and $C$ under a certain linear transformation are known,
then the principle allows us to find the image of $z^*$.

\item[31.] 
On the other hand, if the images of $z$ and $z^*$ are known, we conclude that the image of $C$ must be a line of symmetry of these images.

\item[32.] 
While this is not enough to determine the image of $C$, the information we gain is nevertheless valuable. 

\end{enumerate}

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\begin{frame}{3.3. Symmetry. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[33.] 
The principle of symmetry is put to practical use in the problem of finding the linear transformations which carry a circle $C$ into a circle $C'$.

\item[34.] 
We can always determine the transformation by requiring that three points $z_1$, $z_2$, $z_3$ on $C$ go over into three points $w_1$, $w_2$, $w_3$ on $C'$; the transformation is then $(w,w_1,w_2,w_3) = (z,z_1,z_2,z_3)$.

\end{enumerate}

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\begin{frame}{3.3. Symmetry. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[35.] 
But the transformation is also determined if we prescribe that a point $z_1$ on $C$ shall correspond to a point $w_1$ on $C'$ and that a point $z_2$ not on $C$ shall be carried into a point $w_2$ not on $C'$.

\item[36.] 
We know then that $z_2^*$ (the symmetric point of $z_2$ with respect to $C$) must correspond to $w_2^*$ (the symmetric point of $w_2$ with respect to $C'$). 

\item[37.] 
Hence the transformation will be obtained from the relation 
$$(w,w_1,w_2,w_2^*) = (z,z_1,z_2,z_2^*). $$


\end{enumerate}

\end{frame}

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\begin{frame}{3.3. Symmetry. Exercise 1. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Prove that every reflection carries circles into circles.
}

\item  


\end{itemize}

\end{frame}

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\begin{frame}{3.3. Symmetry. Exercise 2. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Reflect the imaginary axis, the line $x = y$, and the circle $|z| = 1$ in the circle $|z - 2| = 1$.
}

\item  


\end{itemize}

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\begin{frame}{3.3. Symmetry. Exercise 3. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Carry out the reflections in the preceding exercise by geometric construction.
}

\item  


\end{itemize}

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\begin{frame}{3.3. Symmetry. Exercise 4. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Find the linear transformation which carries the circle $|z| = 2$ into $|z + 1| = 1$, the point $-2$ into the origin, and the origin into $i$.
}

\item  



\end{itemize}

\end{frame}

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\begin{frame}{3.3. Symmetry. Exercise 5. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Find the most general linear transformation of the circle $|z| = R$ into itself.
}

\item  


\end{itemize}

\end{frame}

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\begin{frame}{3.3. Symmetry. Exercise 6. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Suppose that a linear transformation carries one pair of concentric circles into another pair of concentric circles. Prove that the ratios of the radii must be the same.
}

\item  


\end{itemize}

\end{frame}

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\begin{frame}{3.3. Symmetry. Exercise 7. \hfill 作业3C-7 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Find a linear transformation which carries $|z| = 1$ and $|z - \frac{1}{4}| = \frac{1}{4}$ into concentric circles. What is the ratio of the radii?
}

\item  



\end{itemize}

\end{frame}

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\begin{frame}{3.3. Symmetry. Exercise 8. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Same problem for $|z| = 1$ and $x = 2$.
}

\item  


\end{itemize}

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\begin{frame}{3.4. Oriented Circles. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.] 
Because $S(z)$ is analytic and
\begin{equation*}
S'(z) = \frac{ad-bc}{(cz+d)^2} \neq 0
\end{equation*}
the mapping $w = S(z)$ is conformal for $z \neq -d/c$ and $\infty$. 

\item[2.] 
It follows that a pair of intersecting circles are mapped on circles that include the same angle. 

\item[3.] 
In addition, {\color{blue}the sense of an angle is preserved}. 

\item[4.] 
From an intuitive point of view this means that right and left are preserved, but a more precise formulation is desirable.


\end{enumerate}

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\begin{frame}{3.4. Oriented Circles. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[5.] 
An orientation of a circle $C$ is determined by an ordered triple of points $z_1,z_2,z_3$ on $C$.

\item[6.] 
{\color{red}With respect to this orientation a point $z$ not on $C$ is said to {\color{blue}lie to the right of $C$} if $\mathrm{Im}\, (z,z_1,z_2,z_3) > 0$ and {\color{blue}to the left of $C$} is $\mathrm{Im}\, (z,z_1,z_2,z_3) < 0$ 
(this checks with everyday use because $(i,1,0, \infty) = i$).
}

\item[7.] 
It is essential to show that there are only two different orientations.

\item[8.] 
By this we mean that the distinction between left and right is the same for all triples, while the meaning may be reversed. 

\item[9.] 
Since the cross ratio is invariant, it is sufficient to consider the case where $C$ is the real axis. 

\end{enumerate}

\end{frame}

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\begin{frame}{3.4. Oriented Circles. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[10.] 
Then
\begin{equation*}
(z,z_1,z_2,z_3) = \frac{az + b}{cz + d}
\end{equation*}
can be written with real coefficients, and a simple calculation gives 
\begin{equation*}
\mathrm{Im}\, (z,z_1,z_2,z_3) = \frac{ad-bc}{|cz+d|^2}\mathrm{Im}\, z.
\end{equation*}

\item[11.] 
We recognize that the distinction between right and left is the same as the distinction between the upper and lower half plane. 

\item[12.] 
Which is which depends on the sign of the determinant $ad-bc$. 

\end{enumerate}

\end{frame}

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\begin{frame}{3.4. Oriented Circles. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[13.] 
A linear transformation $S$ carries the oriented circle $C$ into a circle which we orient through the triple $Sz_1, Sz_2, Sz_3$. 

\item[14.] 
From the invariance of the cross ratio it follows that the left and right of $C$ will be mapped on the left and right of the image circle. 

\end{enumerate}

\end{frame}

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\begin{frame}{3.4. Oriented Circles. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[15.] 
If two circles are tangent to each other, their orientations can be compared.

\item[16.] 
Indeed, we can use a linear transformation which throws their common point to $\infty$. 

\item[17.] 
The circles become parallel straight lines, and we know how to compare the directions of parallel lines. 

\end{enumerate}

\end{frame}

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\begin{frame}{3.4. Oriented Circles. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[18.] 
In the geometric representation the orientation $z_1, z_2, z_3$ can be indicated by an arrow which points from $z_1$ over $z_2$ to $z_3$.

\item[19.] 
With the usual choice of the coordinate system left and right will have their customary meaning with respect to this arrow. 

\end{enumerate}

\end{frame}

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\begin{frame}{3.4. Oriented Circles. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[20.] 
When the finite plane is considered as part of the extended plane, the point at infinity is distinguished. 

\item[21.] 
{\color{red}We can therefore define an {\color{blue}absolute positive orientation} of all finite circles by the requirement that $\infty$ should lie to the right of the oriented circles. }

\item[22.] 
The points to the left are said to form the inside of the circle and the points to the right form its outside.

\end{enumerate}

\end{frame}

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\begin{frame}{3.4. Oriented Circles. Exercise 1. \hfill 作业3C-8 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If $z_1, z_2, z_3, z_4$ are points on a circle, show that $z_1, z_3, z_4$ and $z_2, z_3, z_4$ determine the same orientation if and only if $(z_1,z_2,z_3,z_4) > 0$.
}

\item  


\end{itemize}

\end{frame}

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\begin{frame}{3.4. Oriented Circles. Exercise 2. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Prove that a tangent to a circle is perpendicular to the radius through the point of contact (in this connection a tangent should be defined as a straight line with only one point in common with the circle).
}

\item  


\end{itemize}

\end{frame}

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\begin{frame}{3.4. Oriented Circles. Exercise 3. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Verify that the inside of the circle $|z - a| = R$ is formed by all points $z$ with $|z - a| < R$.
}

\item  


\end{itemize}

\end{frame}

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\begin{frame}{3.4. Oriented Circles. Exercise 4. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
The angle between two oriented circles at a point of intersection is
defined as the angle between the tangents at that point, equipped with the
same orientation. Prove by analytic reasoning, rather than geometric
inspection, that the angles at the two points of intersection are opposite
to each other.

}

\item  


\end{itemize}

\end{frame}


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\begin{frame}{3.5. Families of Circles.  }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.] 
{\color{red}A great deal can be done toward the visualization of linear transformations by the introduction of certain families of circles which may be thought of as {\color{blue}coordinate lines} in a {\color{blue}circular coordinate system}.
}

\item[2.] 
Consider a linear transformation of the form 
\begin{equation*}
w=k\frac{z-a}{z-b}. 
\end{equation*}

\item[3.] 
Here $z = a$ corresponds to $w = 0$ and $z = b$ to $w = \infty$. 

\item[4.] 
It follows that the straight lines through the origin of the $w$-plane are images of the circles through $a$ and $b$.

\end{enumerate}

\end{frame}

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\begin{frame}{3.5. Families of Circles.  }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[5.] 
On the other hand, the concentric circles about the origin, $|w| = \rho$, correspond to circles with the equation
\begin{equation*}
\left\vert \frac{z-a}{z-b} \right\vert = \frac{\rho}{|k|}. 
\end{equation*}

\item[6.] 
{\color{red}These are the circles of {\color{blue}Apollonius} with limit points $a$ and $b$. }

\item[7.] 
By their equation they are the loci of points whose distances from $a$ and $b$ have a constant ratio. 

\item[8.] 
Denote by $C_1$ the circles through $a$, $b$ and by $C_2$ the {\color{blue}circles of Apollonius} with these limit points. 

\item[9.] 
The configuration (Fig. 3-3) formed by all the circles $C_1$ and $C_2$ will be referred to as the {\color{blue}circular net} or the {\color{blue}Steiner circles} determined by $a$ and $b$. 

\end{enumerate}

\end{frame}

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\begin{frame}{3.5. Families of Circles.  }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{figure}[ht!]
\centering
\includegraphics[height=0.75\textheight, width=0.5\textwidth]{figure-3-3.png}
%\caption{Fig. 3-1. Doubly covered region }
\end{figure}

\end{frame}

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\begin{frame}{3.5. Families of Circles.  }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[10.] 
It has many interesting properties.  List a few. 

\begin{enumerate}
\item[1.] 
There is exactly one $C_1$ and one $C_2$ through each point in the plane with the exception of the limit points.

\item[2.] 
Every $C_1$ meets every $C_2$ under right angles.

\item[3.] 
Reflection in a $C_1$ transforms every $C_2$ into itself and every $C_1$ into another $C_1$.
%
Reflection in a $C_2$ transforms every $C_1$ into itself and every $C_2$ into another $C_2$.

\item[4.] 
The limit points are symmetric with respect to each $C_2$, but not with respect to any other circle.

\end{enumerate}

\end{enumerate}

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\begin{frame}{3.5. Families of Circles.  }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[5.] 
These properties are all trivial when the limit points are $0$ and $\infty$, i.e., when the $C_1$ are lines through the origin and the $C_2$ concentric circles.

\item[6.] 
Since the properties are invariant under linear transformations, they must continue to hold in the general case.


\item[7.] 
If a transformation $w = Tz$ carries $a, b$ into $a', b'$ it can be written in the form
\begin{equation}
\frac{w-a'}{w-b'} = k \frac{z-a}{z-b}. 
\label{eq-12}
\end{equation}

\item[8.] 
It is clear that $T$ transforms the circles $C_1$ and $C_2$ into circles $C_1'$ and
$C_2'$ with the limit points $a', b'$.

\end{enumerate}

\end{frame}

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\begin{frame}{3.5. Families of Circles.  }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[9.] 
The situation is particularly simple if $a' = a$, $b' = b$.

\item[10.] 
Then $a, b$ are said to be fixed points of $T$, and it is convenient to represent $z$ and $Tz$ in the same plane. 

\item[11.] 
Under these circumstances the whole circular net will be mapped upon itself. 

\item[12.] 
The value of $k$ serves to identify the image circles $C_1'$ and $C_2'$. 

\item[13.] 
{\color{red}Indeed, with appropriate orientations $C_1$ forms the angle $\mathrm{arg}\, k$ with its image $C_1'$ and the quotient of the constant ratios $|z-a|/|z-b|$ on $C_2'$ and $C_2$ is $|k|$.
}

\end{enumerate}

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\begin{frame}{3.5. Families of Circles.  }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}



\item[14.] 
The special cases in which all $C_1$ or all $C_2$ are mapped upon themselves are particularly important.

\item[15.] 
We have $C_1' = C_1$ for all $C_1$ if $k > 0$ (if $k < 0$ the circles are still the same, but the orientation is reversed). 

\item[16.] 
{\color{red}The transformation is then said to be {\color{blue}hyperbolic}.}

\item[17.] 
When $k$ increases the points $Tz, z \neq a, b$, will flow along the circles $C_1$ toward $b$. 

\item[18.] 
The consideration of this flow provides a very clear picture of a hyperbolic transformation. 

\end{enumerate}

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\begin{frame}{3.5. Families of Circles.  }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[19.] 
The case $C_2' = C_2$ occurs when $|k| = 1$. 

\item[20.] 
Transformations with this property are called elliptic.

\item[21.] 
When $\mathrm{arg}\, k$ varies, the points $Tz$ move along the circles $C_2$.

\item[22.] 
The corresponding flow circulates about $a$ and $b$ in different directions. 


\end{enumerate}

\end{frame}

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\begin{frame}{3.5. Families of Circles.  }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}



\item[23.] 
The general linear transformation with two fixed points is the product of a hyperbolic and an elliptic transformation with the same fixed points. 

\item[24.] 
The fixed points of a linear transformation are found by solving the equation
\begin{equation}
z = \frac{\alpha z+\beta}{\gamma z+\delta}. 
\label{eq-13}
\end{equation}

\item[25.] 
In general this is a quadratic equation with two roots; if $\gamma = 0$ one of the fixed points is $\infty$. 

\item[26.] 
It may happen, however, that the roots coincide.

\item[27.] 
A linear transformation with coinciding fixed points is said to be parabolic.

\end{enumerate}

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\begin{frame}{3.5. Families of Circles.  }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[28.] 
The condition for this is $(\alpha -\delta)^2 = 4\beta\gamma$.

\item[29.] 
If the equation (\ref{eq-13}) is found to have two distinct roots $a$ and $b$, the 
transformation can be written in the form
\begin{equation*}
\frac{w-a}{w-b} = k \frac{z-a}{z-b}. 
%\label{eq-}
\end{equation*}

\item[30.] 
We can then use the Steiner circles determined by $a, b$ to discuss the nature of the transformation. 

\item[31.] 
It is important to note, however, that the method is by no means restricted to this case.

\item[32.] 
We can write any linear transformation in the form (\ref{eq-12}) with arbitrary $a, b$ and use the two circular nets to great advantage.


\end{enumerate}

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\begin{frame}{3.5. Families of Circles.  }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}



\item[33.] 
For the discussion of parabolic transformations it is desirable to introduce still another type of circular net.

\item[34.] 
Consider the transformation
\begin{equation*}
w = \frac{\omega}{z-a} + c. 
%\label{eq-}
\end{equation*}

\item[35.] 
It is evident that straight lines in the $w$-plane correspond to circles through $a$; moreover, parallel lines correspond to mutually tangent circles.

\item[36.] 
In particular, if $w = u + iv$ the lines $u = \mathrm{constant}$ and $v = \mathrm{constant}$ correspond to two families of mutually tangent circles which intersect at right angles (Fig. 3-4). 

\end{enumerate}

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\begin{frame}{3.5. Families of Circles. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}


\begin{figure}[ht!]
\centering
\includegraphics[height=0.75\textheight, width=0.5\textwidth]{figure-3-4.png}
%\caption{Fig. 3-4. Degenerate Steiner circles }
\end{figure}

\end{frame}

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\begin{frame}{3.5. Families of Circles.  }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[37.] 
This configuration can be considered as a degenerate set of Steiner circles. 

\item[38.] 
It is determined by the point $a$ and the tangent to one of the families of circles.

\item[39.] 
We shall denote the images of the lines $v = \mathrm{constant}$ by $C_1$, the circles of the other family by $C_2$. 

\item[40.] 
Clearly, the line $v = \mathrm{Im}\, c$ corresponds to the tangent of the circles $C_1$; its direction is given by $\mathrm{arg}\, \omega$.

\end{enumerate}

\end{frame}

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\begin{frame}{3.5. Families of Circles.  }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[41.] 
Any transformation which carries $a$ into $a'$ can be written in the form 
\begin{equation*}
\frac{\omega'}{w-a'} = \frac{\omega}{z-a} + c. 
%\label{eq-}
\end{equation*}

\item[42.] 
It is clear that the circles $C_1$ and $C_2$ are carried into the circles $C_1'$ and
$C_2'$ determined by $a'$ and $\omega'$.

\item[43.] 
We suppose now that $a = a'$ is the only fixed point. 

\item[44.] 
Then $\omega = \omega'$ and we can write
\begin{equation}
\frac{\omega}{w-a} = \frac{\omega}{z-a} + c. 
\label{eq-14}
\end{equation}

\end{enumerate}

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\begin{enumerate}


\item[45.] 
By this transformation the configuration consisting of the circles $C_1$ and $C_2$ is mapped upon itself.

\item[46.] 
In (\ref{eq-14}) a multiplicative factor is arbitrary, and we can hence suppose that $c$ is real.

\item[47.] 
Then every $C_1$ is mapped upon itself and the parabolic transformation can be considered as a flow along the circles $C_2$.

\item[48.] 
A linear transformation that is neither hyperbolic, elliptic, nor parabolic is said to be {\color{blue}loxodromic}.

\end{enumerate}

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\begin{frame}{3.5. Families of Circles. Exercise 1. }

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\begin{itemize}
\item  {\color{red}Question. 
Find the fixed points of the linear transformations
$$
w=\frac{z}{2z-1}, \hspace{0.5cm}
w=\frac{2z}{3z-1}, \hspace{0.5cm}
w=\frac{3z-4}{z-1}, \hspace{0.5cm}
w=\frac{z}{2-z}. 
$$
Is any of these transformations elliptic, hyperbolic, or parabolic?
}

\item  



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\begin{frame}{3.5. Families of Circles. Exercise 2. }

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\begin{itemize}
\item  {\color{red}Question. 
Suppose that the coefficients of the transformation
$$
Sz = \frac{az + b}{cz + d}
$$
are normalized by $ad - bc = 1$. Show that $S$ is elliptic if and only if 
$-2 <a+d < 2$, parabolic if $a+d =\pm 2$, hyperbolic if $a+d < -2$ or $>2$.
}

\item  


\end{itemize}

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\begin{frame}{3.5. Families of Circles. Exercise 3. }

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\begin{itemize}
\item  {\color{red}Question. 
Show that a linear transformation which satisfies $S^nz = z$ for some integer $n$ is necessarily elliptic.
}

\item  


\end{itemize}

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\begin{frame}{3.5. Families of Circles. Exercise 4. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If $S$ is hyperbolic or loxodromic, show that $S^nz$ converges to a fixed point as $n \to\infty$, the same for all $z$, except when $z$ coincides with the other fixed point. (The limit is the attractive, the other the repellent fixed point. 
What happens when $n \to -\infty$? What happens in the parabolic case?)
}

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\begin{frame}{3.5. Families of Circles. Exercise 5. }

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\begin{itemize}
\item  {\color{red}Question. 
Find all linear transformations which represent rotations of the Riemann sphere.
}

\item  

\end{itemize}

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\begin{frame}{3.5. Families of Circles. Exercise 6. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Find all circles which are orthogonal to $|z| = 1$ and $|z-1| = 4$.
}

\item  


\end{itemize}

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\begin{frame}{3.5. Families of Circles. Exercise 7. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
In an obvious way, which we shall not try to make precise, a family of transformations depends on a certain number of real parameters. 
How many real parameters are there in the family of all linear transformations?
How many in the families of hyperbolic, elliptic, parabolic transformations?
How many linear transformations leave a given circle $C$ invariant?
}

\item  


\end{itemize}

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